Question

我有下面的代码，我期望从嵌套数组映射结果并返回一个具有两个id的单个数组，但我得到2个数组。有人可以指导我做错了什么吗？

arrayVal = [{
    sources: {
      data: [{
        id: 1
      }]
    }
  },
  {
    sources: {
      data: [{
        id: 2
      }]
    }
  }
]

for (let sub of arrayVal) {
  let result = sub.sources.data.map(x => (x.id))
  console.log(result)
}

Answer 1

现在，您正在为map中的每个元素调用arrayVal，因此您将获得两个数组。请改用reduce，将一个对象数组转换为另一个数组，该数组与输入元素不一定是：

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const arrayVal=[{sources:{data:[{id:1}]}},{sources:{data:[{id:2}]}}]; const result = arrayVal.reduce((a, { sources: { data } }) => ( [...a, ...data.map(({ id }) => id)] ), []); console.log(result)
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Answer 2

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arrayVal = [{
    sources: {
      data: [{
        id: 1
      }]
    }
  },
  {
    sources: {
      data: [{
        id: 2
      }]
    }
  }
]

let result = [];
for (let sub of arrayVal) {
  result = result.concat(sub.sources.data.map(x => (x.id)))

}

console.log(result)

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我认为concat就是你在这里所缺少的，希望这是你想要实现的目标

Answer 3

尝试以下

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var arrayVal = [{sources: {data: [{id: 1}]}},{sources: {data: [{id: 2}]}}];

// Create an array on sources.data and merge it into 1 collection (array)
var result = arrayVal.reduce((a, c) => [...a, ...c.sources.data.map(({id}) => id)], []);

console.log(result);

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供参考，Array.reduce

此外，您可以按如下方式改进代码

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var arrayVal = [{sources: {data: [{id: 1}]}},{sources: {data: [{id: 2}]}}];

let result = [];
for (let sub of arrayVal) {
  result.push(sub.sources.data.map(x => (x.id)));
}
console.log([].concat(...result))

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Answer 4

试试这个

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`Route::middleware(['auth', 'moderator|admin'])->group(function() {
   Route::get('/dashboard', function() {
       return view('dashboard');
   });
});`

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Answer 5

您可以这样做：

arrayVal = [{
    sources: {
      data: [{
        id: 1
      }]
    }
  },
  {
    sources: {
      data: [{
        id: 2
      }]
    }
  }
]

var flat = arrayVal.reduce(function(prev,curr,cI){
  prev.push(curr.sources.data.map(x => (x.id))[0]);
  return prev; // *********  Important ******
}, []);

array.map（）不返回单个数组

5 个答案: